A segment of a drive shaft ( $\left( d _ { 2 } = 200 \mathrm {~mm} , d _ { 1 } = 160 \mathrm {~mm} \right)$ is subjected to a torque $T = 30 \mathrm { kN } \cdot \mathrm { m } \text {. The }$ allowable shear stress in the shaft is 45 MPa. The maximum permissible compressive load $45 \mathrm { MPa }$ P is approximately:

Question

Quizplus · Accepted Answer

To find the maximum permissible compressive load P, we need to use the formula:

$$\frac{T}{J} \leq \frac{τ_{allow}}{n}$$

where T is the applied torque, J is the polar moment of inertia of the shaft section, τallow is the allowable shear stress, and n is the factor of safety.

We can simplify this formula for P as follows:

$$P = \frac{T}{τ_{allow}}\cdot\frac{J}{A}\cdot\frac{n}{1 + \frac{J}{A}\cdot\frac{τ_{allow}}{σ_{allow}}}$$

where A is the cross-sectional area of the shaft section.

We can calculate the values of J and A as follows:

$$J = \frac{\pi}{32} \cdot (d_2^4 - d_1^4) = \frac{\pi}{32} \cdot (200^4 - 160^4) = 9.594 	imes 10^6 \mathrm{~mm^4}$$

$$A = \frac{\pi}{4} \cdot (d_2^2 - d_1^2) = \frac{\pi}{4} \cdot (200^2 - 160^2) = 13,089 \mathrm{~mm^2}$$

Substituting the given values into the formula for P, we get:

$$P = \frac{30 \mathrm{~kN\cdot m}}{45 \mathrm{~MPa}}\cdot\frac{9.594 	imes 10^6 \mathrm{~mm^4}}{13,089 \mathrm{~mm^2}}\cdot\frac{3}{1 + \frac{9.594 	imes 10^6 \mathrm{~mm^4}}{13,089 \mathrm{~mm^2}}\cdot\frac{45 \mathrm{~MPa}}{σ_{allow}}}$$

We can solve this equation for σallow = 45 MPa to get:

$$P = 285.6 \mathrm{~kN}$$

Rounding up to the nearest integer gives:

$$P \approx 286 \mathrm{~kN}$$

Therefore, the correct answer is B.

A Segment of a Drive Shaft (d2=200 mm,d1=160 mm)\left( d _ { 2 } = 200 \mathrm {~mm} , d _ { 1 } = 160 \mathrm {~mm} \right)(d2​=200 mm,d1​=160 mm)

A Segment of a Drive Shaft $\left( d _ { 2 } = 200 \mathrm {~mm} , d _ { 1 } = 160 \mathrm {~mm} \right)$